Question #0a956

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Question #0a956

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Answer 1 · 2017-04-07T22:38:22

$141.26; 218.74$ $141.26; 218.74$

Explanation:

cos 2t + 10cos t - 8 = 0
Replace cos 2t by (2cos^2 t - 1):
$2 {cos}^{2} t - 1 + 10 cos t - 8 = 0$ $2cos^2 t - 1 + 10cost - 8 = 0$
Solve this quadratic equation for cos t:
$2 {cos}^{2} t + 10 cos t - 9 = 0$ $2cos^2 t + 10cos t - 9 = 0$ .
$D = b^{2} - 4 a c = 100 + 72 = 172$ $D = b^2 - 4ac = 100 + 72 = 172$ --> $d = \pm 2 \sqrt{43}$ $d = +- 2sqrt43$
There are 2 real roots:
$cos t = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{10}{4} \pm 2 \sqrt{43} \frac{)}{4}$ $cos t = -b/(2a) +- d/(2a) = 10/4 +- 2sqrt43)/4$
cos t = 2.5 + 3.28 = 5.78 (Rejected as > 1) $cos t = 2.5 - 3.28 = - 0.78 Calculator and unit circle give: cos t = - 0.78 --> t = +- 141^@26 Co-terminal arc of (-141.26) -->$ t = (360) - 141^@26 = 218^@74 $Answers for (0, 2pi):$ 141^@26; 218^@74#

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Question #0a956

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